Problem: Captain Daniel has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Kevin and his merciless band of thieves. The Captain has probability $\dfrac{4}{7}$ of hitting the pirate ship, if his ship hasn't already been hit. If it has been hit, he will always miss. The pirate has probability $\dfrac{1}{2}$ of hitting the Captain's ship, if his ship hasn't already been hit. If it has been hit, he will always miss as well. If the pirate shoots first, what is the probability that the pirate hits the Captain's ship, but the Captain misses?
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is the pirate hitting the Captain's ship and event B is the Captain missing the pirate ship. The pirate fires first, so his ship can't be sunk before he fires his cannons. So, the probability of the pirate hitting the Captain's ship is $\dfrac{1}{2}$ If the pirate hit the Captain's ship, the Captain has no chance of firing back. So, the probability of the Captain missing the pirate ship given the pirate hitting the Captain's ship is $1$ The probability that the pirate hits the Captain's ship, but the Captain misses is then the probability of the pirate hitting the Captain's ship times the probability of the Captain missing the pirate ship given the pirate hitting the Captain's ship. This is $\dfrac{1}{2} \cdot 1 = \dfrac{1}{2}$